Croatian Junior Mathematical Olympiad

Let us denote $x = a-1$, $y = b-1$ and $z = c-1$. Then $x, y, z \ge 0$, and the given inequality easily transforms into
$$3xyz + xy + yz + zx \ge 0,$$
which is true since all addends on the left-hand side are non-negative.

The equality is obtained if and only if $xyz = xy = yz = zx = 0$, which is true if and only if at least two numbers among $x, y$ and $z$ are equal to $0$, i.e. if and only if at least two numbers among $a, b$ and $c$ are equal to $1$.